A051386 Numbers whose 4th power is the sum of two positive cubes.

2, 9, 16, 28, 35, 54, 65, 72, 91, 126, 128, 133, 134, 152, 182, 183, 189, 201, 217, 219, 224, 243, 250, 273, 278, 280, 309, 341, 344, 351, 370, 399, 407, 422, 432, 453, 468, 497, 513, 520, 539, 559, 576, 579, 637, 651, 658, 686, 728, 730, 737, 756, 793, 854

Offset: 1

Author

Jud McCranie

Keywords

• nonn

Comments

n such that n^4 = r^3 + s^3 has a solution with r>0, s>0.

By multiplying n^4 = r^3 + s^3 by n^3, also numbers whose 7th power is expressible as the sum of positive cubes.

When n is the sum of 2 positive cubes (A003325) there is a trivial solution: e.g., 133 is a term in A003325, 133=2^3+5^3 and 133^4=(2*133)^3+(5*133)^3. - Zak Seidov, Oct 17 2011

From Robert Israel, Jun 01 2015: (Start)

Slightly more generally, if x^3 + y^3 = u*v^4, then (u*v*w^3)^4 = (u*w^4*x)^3 + (u*w^4*y)^3, so u*v*w^3 is in the sequence for any w >= 1.

There are at least five pairs of adjacent numbers in the sequence: (133,134),(182,183), (854,855), (1842,1843), (3473,3474). Are there infinitely many?

(End)

Examples

			134^4 = 469^3 + 603^3.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A003325, A051387.

Programs

Maple

N:= 1000: # to get all terms <= N
Cubes:= {seq(x^3,x=1..floor(N^(4/3)))}:
select(n -> nops(map(t -> n^4-t, Cubes) intersect Cubes)>0, [$1..N]); # Robert Israel, Jun 01 2015